[proofplan] We use the operator-theoretic definition of the field trace: $\operatorname{Tr}_{K/k}(x)$ is the trace of the $k$-linear multiplication map $m_x: K \to K$. The multiplication map associated to $a\alpha + b\beta$ is exactly the $k$-linear combination $a m_\alpha + b m_\beta$. Linearity of the ordinary trace on the finite-dimensional $k$-[vector space](/page/Vector%20Space) $K$ then gives the desired identity. [/proofplan]

[step:Represent field elements by multiplication endomorphisms] Since $K/k$ is finite, $K$ is a finite-dimensional vector space over $k$. For each $x \in K$, define the multiplication map \begin{align*} m_x: K \to K, \quad y \mapsto xy. \end{align*} This map is $k$-linear: for all $c,d \in k$ and all $y,z \in K$, \begin{align*} m_x(cy + dz) = x(cy + dz). \end{align*} By distributivity in $K$, \begin{align*} x(cy + dz) = cxy + dxz. \end{align*} Since $c,d \in k \subset K$, \begin{align*} cxy + dxz = c m_x(y) + d m_x(z). \end{align*} Let $\operatorname{End}_k(K)$ denote the $k$-vector space of $k$-linear endomorphisms of $K$. Thus $m_x \in \operatorname{End}_k(K)$. By the definition of the field trace through multiplication endomorphisms, \begin{align*} \operatorname{Tr}_{K/k}(x) = \operatorname{tr}_k(m_x), \end{align*} where $\operatorname{tr}_k$ denotes the ordinary trace of a $k$-linear endomorphism of the finite-dimensional $k$-vector space $K$. [/step]

[step:Identify the multiplication map of a linear combination]Let $a,b \in k$ and $\alpha,\beta \in K$. We prove that \begin{align*} m_{a\alpha + b\beta} = a m_\alpha + b m_\beta \end{align*} as elements of $\operatorname{End}_k(K)$. For any $y \in K$, \begin{align*} m_{a\alpha + b\beta}(y) = (a\alpha + b\beta)y. \end{align*} Using distributivity in $K$ and associativity of multiplication, \begin{align*} (a\alpha + b\beta)y = a(\alpha y) + b(\beta y). \end{align*} By the definitions of $m_\alpha$ and $m_\beta$, \begin{align*} a(\alpha y) + b(\beta y) = a m_\alpha(y) + b m_\beta(y). \end{align*} Therefore \begin{align*} m_{a\alpha + b\beta}(y) = a m_\alpha(y) + b m_\beta(y). \end{align*} Since the two $k$-linear maps agree on every $y \in K$, they are equal.[/step]

[guided]The point of this step is to translate the algebraic expression $a\alpha + b\beta$ into a statement about linear maps. We have already defined, for each $x \in K$, the map \begin{align*} m_x: K \to K, \quad y \mapsto xy. \end{align*} Now fix $a,b \in k$ and $\alpha,\beta \in K$. To compare $m_{a\alpha + b\beta}$ with $m_\alpha$ and $m_\beta$, we evaluate both sides on an arbitrary element $y \in K$. By definition of multiplication by $a\alpha + b\beta$, \begin{align*} m_{a\alpha + b\beta}(y) = (a\alpha + b\beta)y. \end{align*} The field multiplication in $K$ is distributive, so \begin{align*} (a\alpha + b\beta)y = (a\alpha)y + (b\beta)y. \end{align*} Associativity of multiplication gives \begin{align*} (a\alpha)y + (b\beta)y = a(\alpha y) + b(\beta y). \end{align*} Since $m_\alpha(y) = \alpha y$ and $m_\beta(y) = \beta y$, this is exactly \begin{align*} m_{a\alpha + b\beta}(y) = a m_\alpha(y) + b m_\beta(y). \end{align*} The right-hand side is the value at $y$ of the $k$-linear endomorphism $a m_\alpha + b m_\beta$. Because this equality holds for every $y \in K$, the two maps are the same element of $\operatorname{End}_k(K)$: \begin{align*} m_{a\alpha + b\beta} = a m_\alpha + b m_\beta. \end{align*}[/guided]

[step:Apply linearity of the ordinary trace] The ordinary trace map \begin{align*} \operatorname{tr}_k: \operatorname{End}_k(K) \to k \end{align*} is $k$-linear on endomorphisms of the finite-dimensional $k$-vector space $K$. Therefore, \begin{align*} \operatorname{tr}_k(m_{a\alpha + b\beta}) = \operatorname{tr}_k(a m_\alpha + b m_\beta). \end{align*} By linearity of $\operatorname{tr}_k$, \begin{align*} \operatorname{tr}_k(a m_\alpha + b m_\beta) = a\operatorname{tr}_k(m_\alpha) + b\operatorname{tr}_k(m_\beta). \end{align*} Using the definition of the field trace for $a\alpha + b\beta$, $\alpha$, and $\beta$, \begin{align*} \operatorname{Tr}_{K/k}(a\alpha + b\beta) = a\operatorname{Tr}_{K/k}(\alpha) + b\operatorname{Tr}_{K/k}(\beta). \end{align*} This proves that $\operatorname{Tr}_{K/k}: K \to k$ is $k$-linear. [/step]